# Determine the empirical and molecular formula of the unknown acid?

## Combustion of 19.81 mg of an industrial acid produces 41.98 mg of CO2 and 6.45 mg of H2O. If 0.250 moles of the acid has a mass of 41.5 g, determine the empirical and molecular formula of the acid.

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Oct 18, 2017

phtalic acid: ${C}_{6} {H}_{4} {\left(C O O H\right)}_{2}$

#### Explanation:

combustion of 19.81 mg of an industrial acid produces 41.98 mg of CO2 and 6.45 mg of H2O. If 0.250 moles of the acid has a mass of 41.5 g, determine the empirical and molecular formula of the acid.
The acid is an organic acid made of C, H and O with probably formula${C}_{n} {H}_{m} {O}_{2}$

the acid has a molecular mass of $\frac{41.5 g}{0.25 m o l} = 166 \frac{g}{m o l}$ . dividing 19,81 mg for 166g/mol you obtain 0,119 mmol
If You divide 41.98 mg Di $C {O}_{2}$ for 44g/mol that is its MM you obtain the mmol of $C {O}_{2}$ svilupped from the copmbustion =0.954 mmol.

Making tha same procediment for water, you obtain:mmol water =$\frac{6.45}{18} = 0.358$ mmol of water
Now we divide the three numbers for the smaller one.

By dividing 0.954/0.119 you obtain 8 that is 8 mol of $C {O}_{2}$ for 1 mol of acid, By dividing 0.358/0.119 You obtain 3 that is 3 mol of ${H}_{2} O$ produced by the reaction that means 8 atoms of carbon (MA= 6X12= 96 and 6 of Hydrogen (MA= 6 xx 1 =6) in the initial organic acid. Oxygen will be ((166-96-6) =64 g that divided for 16 g/mol(MA O) givew 4 atoms of O in the molecula that will have the formula ${C}_{8} {O}_{4} {H}_{6}$.

It comes to my mind only the phtalic acid (two groups COOH attacked to one benzenic ring) with this formula: ${C}_{6} {H}_{4} {\left(C O O H\right)}_{2}$ .

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